# Calculation of the heating system of a private house: rules and examples of calculation

Heating a private house is a necessary element of comfortable housing. Agree that the arrangement of the heating complex should be approached carefully, as mistakes are expensive. But you have never done such calculations and do not know how to perform them correctly?

We will help you - in our article we will consider in detail how the calculation of the heating system of a private house is done to effectively compensate for heat loss in the winter months.

We give specific examples, supplementing the material of the article with visual photos and useful video tips, as well as relevant tables with indicators and coefficients necessary for calculations.

The content of the article:

## Heat loss of a private house

The building loses heat due to the difference in air temperature inside and outside the house. Heat loss is higher, the more significant the area of the building envelope (windows, roofs, walls, foundations).

Also heat loss connected with the materials of the enclosing structures and their sizes. For example, the heat loss of thin walls is greater than thick.

Effective heating calculation for a private house must take into account the materials used in the construction of building envelopes.

For example, with an equal thickness of a wall made of wood and brick, heat is carried out with different intensities - heat loss through wooden structures is slower. Some materials let heat pass better (metal, brick, concrete), others worse (wood, mineral wool, polystyrene foam).

The atmosphere inside a residential building is indirectly related to the external air environment. Walls, openings of windows and doors, roof and foundation in winter transfer heat from the house to the outside, supplying cold in return. They account for 70-90% of the total heat loss of the cottage.

A constant leak of thermal energy during the heating season also occurs through ventilation and sewage.

When calculating the heat loss of an individual housing construction, these data are usually not taken into account. But the inclusion of heat losses through the sewer and ventilation systems in the general thermal calculation of the house is still the right decision.

It is impossible to calculate the autonomous heating circuit of a country house without evaluating the heat loss of its enclosing structures. More precisely, it will not work determine the power of the boilersufficient to heat the cottage in the most severe frosts.

Analysis of the actual consumption of thermal energy through the walls will allow you to compare the costs of boiler equipment and fuel with the costs of thermal insulation of walling.

After all, the more energy-efficient the house, i.e. the less heat it loses during the winter months, the lower the cost of acquiring fuel.

For a competent calculation of the heating system, you will need coefficient of thermal conductivity common building materials.

### Calculation of heat loss through walls

Using the conditional two-story cottage as an example, we calculate the heat loss through its wall structures.

Initial data:

- square “box” with front walls 12 m wide and 7 m high;
- within the walls of 16 openings, the area of each 2.5 m
^{2}; - material of front walls - full-bodied ceramic brick;
- wall thickness - 2 bricks.

Next, we will calculate the group of indicators from which the total value of heat loss through the walls is added.

#### Heat transfer resistance

To find out the heat transfer resistance index for a facade wall, it is necessary to divide the thickness of the wall material by its heat conductivity coefficient.

For a number of structural materials, data on the coefficient of thermal conductivity are presented in the images above and below.

Our conditional wall is built of solid ceramic brick, the thermal conductivity of which is 0.56 W / m^{about}C. Its thickness, taking into account the masonry on the central distribution center, is 0.51 m. Dividing the wall thickness by the brick thermal conductivity coefficient, we obtain the wall heat transfer resistance:

**0.51: 0.56 = 0.91 W / m ^{2 × o}WITH**

We round the result of the division to two decimal places; there is no need for more accurate data on heat transfer resistance.

#### External Wall Area

Since a square building was chosen as an example, the area of its walls is determined by multiplying the width by the height of one wall, then by the number of external walls:

**12 · 7 · 4 = 336 m ^{2}**

So, we know the area of the front walls. But what about the openings of windows and doors, occupying together 40 m2 (2.5 · 16 = 40 m^{2}) of the front wall, should they be taken into account?

Indeed, how to correctly calculate autonomous heating in a wooden house excluding heat transfer resistance of window and door structures.

If it is necessary to calculate the heat loss of a large-area building or a warm house (energy efficient) - yes, taking into account the heat transfer coefficients of window frames and entrance doors will be correct in the calculation.

However, for low-rise buildings IZHS built from traditional materials, door and window openings can be neglected. Those. Do not take away their area from the total area of the front walls.

#### Common wall heat loss

We find out the heat loss of the wall from its one square meter when the temperature difference between the air inside and outside the house is one degree.

To do this, divide the unit by the heat transfer resistance of the wall, calculated earlier:

**1: 0.91 = 1.09 W / m ^{2}·^{about}WITH**

Knowing the heat loss per square meter of the perimeter of the external walls, you can determine the heat loss at certain street temperatures.

For example, if the temperature in the cottage is +20 ^{about}C, and on the street -17 ^{about}C, the temperature difference will be 20 + 17 = 37 ^{about}C. In this situation, the total heat loss of the walls of our conditional home will be:

**0.91 · 336 · 37 = 11313 W**,

Where: 0.91 - heat transfer resistance per square meter of the wall; 336 - area of front walls; 37 - temperature difference between indoor and outdoor atmosphere.

We recalculate the resulting heat loss in kilowatt hours, they are more convenient for perception and subsequent calculations of the power of the heating system.

#### Wall heat loss in kilowatt hours

First, find out how much thermal energy will go through the walls in one hour with a temperature difference of 37 ^{about}WITH.

We remind you that the calculation is carried out for a house with structural characteristics, conditionally selected for demonstration and demonstration calculations:

**113131: 1000 = 11.313 kWh**,

Where: 11313 - the amount of heat loss obtained earlier; 1 hour; 1000 is the number of watts per kilowatt.

To calculate the heat loss per day, the obtained heat loss per hour is multiplied by 24 hours:

**11.31324 = 271.512 kWh**

For clarity, we find out the loss of thermal energy for the full heating season:

**7 · 30 · 271.512 = 57017.52 kWh**,

Where: 7 - the number of months in the heating season; 30 - the number of days in a month; 271,512 - daily heat loss of the walls.

So, the estimated heat loss of the house with the above characteristics of the building envelope will amount to 57017.52 kWh for seven months of the heating season.

### Taking into account the effects of private house ventilation

As an example, we will calculate the ventilation heat loss during the heating season for a conditional cottage of a square shape, with a wall of 12 meters wide and 7 meters high.

Excluding furniture and interior walls, the internal volume of the atmosphere in this building will be:

**12 · 12 · 7 = 1008 m ^{3}**

At air temperature +20 ^{about}C (norm in the heating season) its density is 1.2047 kg / m^{3}and the specific heat is 1.005 kJ / (kg^{about}WITH).

We calculate the mass of the atmosphere in the house:

**10081.2047 = 1214.34 kg**,

Where: 1008 - the volume of the home atmosphere; 1.2047 - air density at t +20 ^{about}WITH .

Suppose a five-fold change in air volume in the premises of the house. Note that the exact supply volume requirement fresh air depends on the number of residents of the cottage.

With an average temperature difference between the house and the street in the heating season, equal to 27 ^{about}C (20 ^{about}C home, -7 ^{about}With the external atmosphere) per day for heating the supply of cold air you need thermal energy:

**5.271214.34-1.005 = 164755.58 kJ**,

Where: 5 - the number of air changes in the premises; 27 - temperature difference between indoor and outdoor atmosphere; 1214.34 - air density at t +20 ^{about}WITH; 1.005 - specific heat of air.

We convert kilojoules into kilowatt hours, dividing the value by the number of kilojoules in one kilowatt hour (3600):

**164755.58: 3600 = 45.76 kWh**

Having found out the cost of thermal energy for heating the air in the house during its five-fold replacement through the supply ventilation, we can calculate the “air” heat loss for the seven-month heating season:

**7 · 30 · 45.76 = 9609.6 kWh**,

Where: 7 - the number of "heated" months; 30 - the average number of days in a month; 45.76 - daily heat energy costs for heating the supply air.

Ventilation (infiltration) energy consumption is inevitable, since air renewal in the cottage is vital.

The heating needs of the replaceable air atmosphere in the house must be calculated, summed with heat losses through the building envelope and taken into account when choosing a heating boiler. There is another type of heat energy consumption, the latter - sewer heat loss.

### Energy costs for DHW preparation

If during the warmer months cold water flows from the tap into the cottage, then in the heating season it is icy, with a temperature not exceeding +5 ^{about}C. Bathing, washing dishes and washing are not possible without heating the water.

The water drawn into the toilet bowl contacts the home atmosphere through the walls, taking a little heat. What happens to water heated by burning non-free fuel and spent on household needs? It is poured into the sewer.

Let's look at an example. A family of three, suppose to spend 17 m^{3} water monthly. 1000 kg / m^{3} - the density of water, and 4.183 kJ / kg^{about}C is its specific heat.

The average temperature of heating water intended for domestic needs, let it be +40 ^{about}C. Accordingly, the difference in average temperature between cold water entering the house (+5 ^{about}C) and heated in a boiler (+30 ^{about}C) it turns out 25 ^{about}WITH.

To calculate sewer heat loss, we consider:

**17 · 1000 · 25 · 4.183 = 1777775 kJ**,

Where: 17 - monthly volume of water consumption; 1000 is the density of water; 25 - temperature difference between cold and heated water; 4,183 - specific heat of water;

To convert kilojoules to more understandable kilowatt hours:

**1777775: 3600 = 493.82 kWh**

Thus, for a seven-month period of the heating season, heat energy goes into the sewer in the amount of:

**493.827 = 3456.74 kWh**

The consumption of thermal energy for heating water for hygiene needs is small, in comparison with heat loss through walls and ventilation. But this is also energy consumption, loading the boiler or boiler and causing fuel consumption.

## Calculation of the power of the boiler

The boiler in the heating system is designed to compensate for the heat loss of the building. And also, in the case of dual circuit system or when equipping the boiler with an indirect heating boiler, for heating water for hygienic needs.

By calculating the daily heat loss and the consumption of warm water “for sewage”, it is possible to accurately determine the necessary boiler capacity for a cottage of a certain area and the characteristics of the enclosing structures.

To determine the power of the heating boiler, it is necessary to calculate the cost of thermal energy of the house through the facade walls and the heating of the replaceable air atmosphere of the interior.

Data on heat losses in kilowatt hours per day is required - in the case of a conditional house, calculated as an example, this is:

**271.512 + 45.76 = 317.272 kWh**,

Where: 271,512 - daily heat loss by external walls; 45.76 - daily heat loss for heating the supply air.

Accordingly, the necessary heating capacity of the boiler will be:

**317.272: 24 (hours) = 13.22 kW**

However, such a boiler will be under constantly high load, reducing its service life. And on especially frosty days, the rated capacity of the boiler will not be enough, because with a high temperature difference between indoor and outdoor atmospheres, the heat loss of the building will increase sharply.

therefore choose a boiler according to an average calculation of the cost of thermal energy is not worth it - it may not cope with severe frosts.

It will be rational to increase the required capacity of boiler equipment by 20%:

**13.22.2 + 13.22 = 15.86 kW**

To calculate the required power of the second circuit of the boiler, heating water for washing dishes, bathing, etc., it is necessary to divide the monthly heat consumption of the “sewer” heat losses by the number of days in the month and by 24 hours:

**493.82: 30: 24 = 0.68 kW**

According to the calculation results, the optimal boiler power for the cottage example is 15.86 kW for the heating circuit and 0.68 kW for the heating circuit.

## The choice of radiators

Traditionally heating radiator power It is recommended to choose the area of the heated room, with a 15-20% overstatement of power requirements, just in case.

As an example, let us consider how correct the method of choosing a radiator is “10 m2 of area - 1.2 kW”.

Initial data: corner room on the first level of a two-story house IZHS; external wall of double-row ceramic brick masonry; room width 3 m, length 4 m, ceiling height 3 m.

According to the simplified selection scheme, it is proposed to calculate the area of the room, we consider:

**3 (width) · 4 (length) = 12 m ^{2}**

Those. the required power of the heating radiator with a 20% premium is 14.4 kW. Now let's calculate the power parameters of the heating radiator based on the heat loss of the room.

In fact, the area of a room affects the loss of thermal energy less than the area of its walls that extend on one side of the building (front).

Therefore, we will consider exactly the area of "street" walls available in the room:

**3 (width) · 3 (height) + 4 (length) · 3 (height) = 21 m ^{2}**

Knowing the area of the walls that transfer heat “to the street”, we calculate the heat loss with a difference in room and street temperature of 30^{about} (in the house +18 ^{about}C, outside -12 ^{about}C), and immediately in kilowatt hours:

**0.91 · 21 · 30: 1000 = 0.57 kW**,

Where: 0.91 - heat transfer resistance m2 of room walls facing "the street"; 21 - the area of "street" walls; 30 - temperature difference inside and outside the house; 1000 is the number of watts per kilowatt.

It turns out that to compensate for heat loss through the facade walls of this design, at 30^{about} the temperature difference in the house and on the street is enough heating with a capacity of 0.57 kWh. We increase the required power by 20, even by 30% - we get 0.74 kWh.

Thus, the real power requirements of heating can be significantly lower than the trading scheme “1.2 kW per square meter of floor space”.

Moreover, the correct calculation of the required power of heating radiators will reduce the volume coolant in the heating system, which will reduce the load on the boiler and fuel costs.

## Conclusions and useful video on the topic

Where the heat goes from home - the video provides the answers:

In the video, the procedure for calculating the heat loss of a house through the building envelope is considered.Knowing the heat loss, it will be possible to accurately calculate the power of the heating system:

For a detailed video on the principles of selecting the power characteristics of a heating boiler, see below:

Heat production rises annually - fuel prices are rising. And the heat is constantly not enough. You can’t be indifferent to the energy consumption of the cottage - it is completely unprofitable.

**On the one hand, each new heating season costs the homeowner more and more expensive. On the other hand, insulation of walls, foundations and suburban roofs costs good money. However, the less heat that leaves the building, the cheaper it will be to heat it.**.

The preservation of heat in the premises of the house is the main task of the heating system in the winter months. The choice of power of the heating boiler depends on the condition of the house and on the quality of insulation of its enclosing structures. The principle of “kilowatts per 10 squares of area” works in a cottage of an average state of facades, roofs and foundations.

Have you independently calculated a heating system for your home? Or did you notice a mismatch in the calculations given in the article? Share your practical experience or the volume of theoretical knowledge by leaving a comment in the block under this article.

It is good that now there is the opportunity to properly calculate the heating system of a private house. Avoiding mistakes even at the planning stage, we save a lot of money, time and nerves, while getting comfortable living conditions. Previously, everything was done by eye, on a hunch, and often then had to finish or even redo it. It is remarkable that science finds applied application.

If you want a warm and comfortable house, then you need to count on a heating system. Fortunately, there are tons of online calculators on the Internet that simplify the task.